Inverse of [-2, -1, 2; -1, 0, -2; 1, 0, 1] Step by Step

Step by step

The question

Question: find the inverse of A.

A = [−2, −1, 2; −1, 0, −2; 1, 0, 1]

The inverse A⁻¹ is the matrix that times A gives I: 1s down the diagonal, 0s elsewhere.

A × A⁻¹ = I

1 Find the determinant

First we need D, the determinant of A, by the diagonals.

D = |−2, −1, 2; −1, 0, −2; 1, 0, 1|−2, −1; −1, 0; 1, 0

We write the first two columns again, on the right.

Then every diagonal has three numbers.

Down: −2 × 0 × 1, −1 × (−2) × 1, 2 × (−1) × 0.

down = −2 × 0 × 1 + (−1) × −2 × 1 + 2 × −1 × 0

0, 2, 0.

Together: 2.

= 0 + 2 + 0 = 2

Up: 1 × 0 × 2, 0 × (−2) × (−2), 1 × (−1) × (−1).

up = 1 × 0 × 2 + 0 × −2 × (−2) + 1 × −1 × (−1)

0, 0, 1.

Together: 1.

= 0 + 0 + 1 = 1

Down minus up: 2 minus 1 is 1.

So D = 1.

D = 2 − 1 = 1

2 Find the minors

Each entry has a minor: cover its row and its column.

A 2 by 2 is left.

Its determinant, down minus up, is the minor.

minors = [0×1 − 0×(−2), −1×1 − 1×(−2), −1×0 − 1×0; −1×1 − 0×2, −2×1 − 1×2, −2×0 − 1×(−1); −1×(−2) − 0×2, −2×(−2) − −1×2, −2×0 − −1×(−1)]

Work out each one: 0, 1, 0; −1, −4, 1; 2, 6, −1.

= [0, 1, 0; −1, −4, 1; 2, 6, −1]

3 Change the signs

The signs go plus, minus, plus, like a checkerboard.

Where there is a minus, the minor changes sign.

signs: [+, −, +; −, +, −; +, −, +]

So the cofactors are rows 0, −1, 0, 1, −4, −1 and 2, −6, −1.

The purple ones changed sign.

C = [0, −1, 0; 1, −4, −1; 2, −6, −1]

Careful!

Only the minus places change sign: where the row number plus the column number is odd.

Careful: the − places: row + column is odd

4 Swap rows and columns

Row 1 of C becomes column 1, row 2 becomes column 2, row 3 becomes column 3.

Then we divide by D: times 1/1.

A⁻¹ = 11 × [0, 1, 2; −1, −4, −6; 0, −1, −1]

5 Divide by the determinant

Dividing by 1 changes nothing.

So the answer is rows 0, 1, 2, −1, −4, −6 and 0, −1, −1.

A⁻¹ = [0, 1, 2; −1, −4, −6; 0, −1, −1]

6 Check

Let's check: A times our answer must be I.

Row 1 of A times column 1 of the answer: 1.

row 1 × column 1: −2 × 0 + −1 × (−1) + 2 × 0 = 1

Row 1 times column 2: 0.

row 1 × column 2: −2 × 1 + −1 × (−4) + 2 × (−1) = 0

The other entries work the same way: 1s down the diagonal, 0s elsewhere.

So the answer is right!

A × A⁻¹ = I

Let's look back.

First: find the determinant.

Then: find the minors.

Then: change the signs.

Then: swap rows and columns.

Then: divide by the determinant.

Last: check.

Answer